Bakery CodeForces

Masha wants to open her own bakery and bake muffins in one of the n cities numbered from 1 to n. There are m bidirectional roads, each of whose connects some pair of cities.

To bake muffins in her bakery, Masha needs to establish flour supply from some storage. There are only k storages, located in different cities numbereda₁, a₂, ..., a_k.

Unforunately the law of the country Masha lives in prohibits opening bakery in any of the cities which has storage located in it. She can open it only in one of another n - k cities, and, of course, flour delivery should be paid — for every kilometer of path between storage and bakery Masha should pay 1 ruble.

Formally, Masha will pay x roubles, if she will open the bakery in some city b (a_i ≠ b for every 1 ≤ i ≤ k) and choose a storage in some city s (s = a_j for some 1 ≤ j ≤ k) and b and s are connected by some path of roads of summary length x (if there are more than one path, Masha is able to choose which of them should be used).

Masha is very thrifty and rational. She is interested in a city, where she can open her bakery (and choose one of k storages and one of the paths between city with bakery and city with storage) and pay minimum possible amount of rubles for flour delivery. Please help Masha find this amount.

Input

The first line of the input contains three integers n, m and k (1 ≤ n, m ≤ 10⁵, 0 ≤ k ≤ n) — the number of cities in country Masha lives in, the number of roads between them and the number of flour storages respectively.

Then m lines follow. Each of them contains three integers u, v and l (1 ≤ u, v ≤ n,1 ≤ l ≤ 10⁹, u ≠ v) meaning that there is a road between cities u and v of length ofl kilometers .

If k > 0, then the last line of the input contains k distinct integers a₁, a₂, ..., a_k(1 ≤ a_i ≤ n) — the number of cities having flour storage located in. If k = 0 then this line is not presented in the input.

Output

Print the minimum possible amount of rubles Masha should pay for flour delivery in the only line.

If the bakery can not be opened (while satisfying conditions) in any of the ncities, print - 1 in the only line.

Example

Input

Output

Input

3 1 1
1 2 3
3

Output

-1

Note

$\text{[math]}$

Image illustrates the first sample case. Cities with storage located in and the road representing the answer are darkened.

//这一道题之前没有做出来的原因是不知道怎么处理最后输入的那一组数据
//虽然之前做过图论题,但是这个类型的还是第一次做。
//很显然,最后输入的那一组数据,这些点肯定在那n个点中,所以这是从指定点找到其他点的最短路径
//那么这个指定点肯定是要特殊处理的点,前面已经给出了各个点之间的关系,所以要把最后输入的
//特殊的点标记一下,而其他点不是特殊点,所以不用标记,然后从与这些特殊点有关的点的边中找到最短的
//就是最后的答案,最后用一个for循环处理一下这m条路就行了,每次取特殊点到与其相连的点的最短路径
#include<queue>
#include<stack>
#include<vector>
#include<math.h>
#include<stdio.h>
#include<numeric>//STL数值算法头文件
#include<stdlib.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#include<functional>//模板类头文件
using namespace std;

const int INF=0x3f3f3f3f;
const int maxn=500100;

int n,m,k,b;
int vis[maxn];
struct node
{
    int u,v,l;
} a[maxn];

int main()
{
    while(~scanf("%d %d %d",&n,&m,&k))
    {
        int i;
        memset(vis,0,sizeof(vis));
        for(i=0; i<m; i++)
        {
            scanf("%d %d %d",&a[i].u,&a[i].v,&a[i].l);
        }
        for(i=0; i<k; i++)
        {
            scanf("%d",&b);
            vis[b]=1;
        }
        int ans=INF;
        for(i=0; i<m; i++)
        {
            if(vis[a[i].u]&&!vis[a[i].v]) ans=min(ans,a[i].l);
            if(!vis[a[i].u]&&vis[a[i].v]) ans=min(ans,a[i].l);
        }
        if(ans==INF) printf("-1
");
        else  printf("%d
",ans);
    }
    return 0;
}